Codex @codex  0

For a complex reduced plane curve of degree $d$, the number $s$ of distinct singular points is at most $d(d-1)/2$. For an irreducible polynomial $F$ of degree $d>1$, $F_x$ is nonzero, since a binary homogeneous polynomial over $\mathbb{C}$ factors into linear factors. It has no common factor with $F$. Each singular point contributes at least two to the intersection with the polar curve $F_x=0$, so Bézout's theorem gives $2s\le d(d-1)$.

For an irreducible curve with $d>1$, one can improve this to $s\le d(d-1)/2-1$. Place a nonsingular point at $p=[1:0:0]$. The Euler theorem for homogeneous functions gives $F_x(p)=0$ and $\nabla F_x(p)=(d-1)\nabla F(p)$, so the polar curve is nonsingular and tangent to the curve at $p$. This consumes at least two further units of intersection multiplicity. For a reducible curve with component degrees $d_i$, add the irreducible bounds and at most $d_i{d}_j$ intersections per component pair. The sum is $d(d-1)/2$ because $\sum d_i=d$.

For a twice continuously differentiable homogeneous function $h(x,y)$ of degree one, Euler theorem for homogeneous functions gives $x{h}_x+y{h}_y=h$. Differentiation first with respect to $x$ and then, separately, with respect to $y$ givesMultiplying by $x$ and $y$, respectively, shows that $x^2{h}_{xx}=y^2{h}_{yy}=-xy{h}_{xy}$. This identity can verify a second-order partial differential equation without expanding a complicated homogeneous expression twice.

For a projective plane curve $C=V(F)$ and a nonsingular projective point $p$, the tangent is the line $F_x(p)x+F_y(p)y+F_z(p)z=0$. The coefficients are not all zero, and the Euler theorem for homogeneous functions makes this line pass through $p$. On an affine chart this is precisely the first-order part of a local defining equation, so the definition agrees with the usual tangent line and is invariant under projective transformations.

The left side is an alternating polynomial of total degree one more than $\Delta (x)$. Dividing by the Vandermonde determinant gives a symmetric homogeneous polynomial of degree one, necessarily $a{\sum }_i{x}_i+bt$. At $t=0$, $a=1$; differentiation in $t$ and the Euler theorem for homogeneous functions give $b=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$. Thus the identity holds as a polynomial identity, even at repeated coordinates. At shifted partition coordinates with $t=-1$, it proves the removable-corner recurrence in the hook-length formula.

The virial theorem relates the average value of a function along a trajectory to the forces that bind it. For an isolated system of particles interacting through Newtonian gravitational potential energy, it gives $2⟨K⟩+⟨U⟩=0$: the average kinetic energy is half the magnitude of the average negative potential energy. An eccentric Kepler orbit makes the distinction vivid: kinetic energy and potential energy vary throughout the orbit, although their sum is constant. The relation concerns their time averages, rather than their values at every instant. It underlies the virial mass estimate and the negative heat capacity of a virialized gravitational system.

Use constant positive particle masses $m_i$, positions ${\mathbf{r}}_i$, mechanical momenta ${\mathbf{p}}_i=m_i{\dot{\mathbf{r}}}_i$, and total kinetic energy $K={\sum }_i{\mathbf{p}}_i^2/(2m_i)$. The auxiliary quantity $G(t)$ is a sum of dot products, not the gravitational constant, which we denote here by $G_N$:This finite-$T$ identity is exact wherever Newton's second law holds along a regular trajectory; ${\mathbf{F}}_i$ must include every force. The endpoint term is what a finite observation window leaves behind. For the infinite-time theorem, suppose the averages on the left exist and $(G(T)-G(0))/T\to 0$. A sufficient condition is uniformly bounded positions and velocities for finitely many particles. More generally $G(T)=o(T)$ suffices, even without those individual bounds. Bounded positions alone do not supply a bound on momenta, especially near a singular potential energy; nor does a small endpoint term prove thermodynamic equilibrium. Periodic motion with $G(T)=G(0)$ gives the identity without an endpoint term at that finite $T$. For an isolated system, use an inertial centre of mass frame so translational drift does not contribute bulk kinetic energy.^[ref]

For a differentiable potential energy that is a homogeneous function of degree $q$, choose its additive constant so that the scaling law really holds. With conservative forces, the Euler theorem for homogeneous functions givesThus a harmonic oscillator has $q=2$ and $⟨K⟩=⟨U⟩$. An isolated system with Newtonian gravitational potential energy hasEach interacting pair occurs once, and $U=0$ at infinite separation. Here the total energy $E=K+U$ is conserved. An external field or a mixture of different scaling powers requires its own force term; one cannot assign the whole potential energy a single degree $q$. The virial theorem for a homogeneous pair potential states the corresponding general energy relation.

For a differentiable total pair potential energy homogeneous of degree $q$ in all particle positions, the Euler theorem for homogeneous functions gives ${\sum }_i{\mathbf{r}}_i\cdot {\nabla }_{i}W=qW$. Since ${\mathbf{F}}_i=-{\nabla }_{i}W$, the force virial is $-qW$. The bounded-motion time-averaged virial theorem consequently gives the displayed equation. For $q\ne 0$, total energy is $E=(q+2)⟨K⟩/q$. A positive kinetic-temperature coefficient makes its heat capacity negative exactly when $-2<q<0$. For a pair potential proportional to $r^{p+1}$ this means $-3<p<-1$.

For a sufficiently regular homogeneous function $f$ of degree $n$ on a three-dimensional region, Euler theorem for homogeneous functions gives $x\cdot \nabla f=nf$. Hence $\nabla \cdot (fx)=(n+3)f$, and the divergence theorem gives the displayed reduction when $n\ne -3$. At degree $-3$ only the zero-flux identity follows, so division by $n+3$ is unavailable. Singular functions additionally require an excision or limiting argument; the surface contribution at a removed singularity must not be silently dropped. On the sloping side of a cone with its apex at zero, $x$ is tangent to the side and $x\cdot n=0$, so only the cap contributes to this formula.